The two separate waves will arrive at a point with the same phase, and hence undergo constructive interference, if and only if this path difference is equal to any integer value of the wavelength, i.e. $(AB+BC) - (AC') = n\lambda $

What I don't understand is the "will arrive at a point with the same phase" part. Aren't the points $C$ and $C'$ separated in space, by a distance of roughly the same order as $d$? To constructively interfere, these two rays ($AC'$ and $BC$) must continue on to some detector, and somehow meet at the same point in space. How does that happen?

$\begingroup$The diagram is for illustrative purposes of the interference condition only. The actual diffraction requires many layers of the crystal to produce a sharp angle dependence. Beyond that the finite sample size and the finite aperture size of the x-ray source have to be compensated for. See e.g. web.stanford.edu/group/glam/xlab/MatSci162_172/LectureNotes/… for actual instrument geometries.$\endgroup$
– CuriousOneOct 6 '14 at 6:51

$\begingroup$Try drawing two incoming rays such that line $AC'$ from one ray is coincident with line $BC$ from the other, and keep in mind all rays are from an incident parallel wavefront.$\endgroup$
– Carl WitthoftOct 6 '14 at 13:17

$\begingroup$@CarlWitthoft : Do you mean something like this? In that image, the coincident rays (which will interfere at a single point at the detector) come from atoms one down and to the left of each other. But I still have the question - why are virtually all other illustrations of Bragg's law like the one in my post, showing only the wavefront (CC') of the scattered wave. To be detected at the detector, wouldn't this (plane wave) wavefront need to be focused by, say, some lens, to a single point?$\endgroup$
– user114806Oct 7 '14 at 0:27

2 Answers
2

Because of the beam divergence in reality the beam is not 1D like what you are seeing in the image. Having enough distance from the surface, their beam profiles will have overlap and we would see the interference. The amount of overlap and details depend on the beam profile of our source or sources. This divergence is not just for electromagnetic waves, matter-waves like electrons also have the beam divergence. The periodic nature of the scattering medium can give rise to different reflection peaks and changing the interference pattern as well.

You can imagine that the two rays converge at the detector. So they are not parallel but rather the sides of a triangle with base AB and the vertex at the detector. Now realize that AB is of the order of angstroms and the sides (distance to detector) are of the order of tens of centimeters. Now in this triangle draw CC'at a distance of a few angstroms from the base. Remember that the vertex is about 10^9 angstroms from the base). How would the segments AC' and BC will look like? Technically not exactly parallel to each other. But would you be able to "see" that they are not parallel? How far would be the angle between them from zero degrees?

A similar "effect" happens with the rays of light from the Sun or a distant star. They are considered parallel but they are not exactly so. How divergent are they?